I am trying to prove the following:
Suppose $\left(X,\|\cdot\|_{1}\right)$ and $\left(X,\|\cdot\|_{2}\right)$ are Banach spaces over $\mathbb{F}$. Prove that if $\left\|x_{j}\right\|_{1} \rightarrow 0$ implies $\left\|x_{j}\right\|_{2} \rightarrow 0$ then there are two constants $c_{1}, c_{2}>0$ such that $c_{1}\|x\|_{1} \leq\|x\|_{2} \leq c_{2}\|x\|_{1}$ for all $x \in X$.
If I have one of the inequalities, the inverse operator theorem gives me the other one, but I am not sure how to get one of those, I saw a post that shows that if a sequence is Cauchy in one norm implies that is Cauchy in the other norm, that implies that the norms are equivalent, as I am in Banach Spaces, this are complete and therefore Cauchy sequences converge, for example $x_n\to x \in X$, I can consider maybe $\hat{x}_n=(x_n-x)\to 0$ and somehow use the result from that post. Am I in a good way? I am not sure how to conclude here. Or if I am wrong about this idea can you give me a hint please? Thanks.
There is no completeness involved in this result so there is no point in looking at Cauchy sequences.
A linear map $T$ from one normed linear space to another is continuous if and only if it is continuous at $0$ iff $\|Tx\| \leq C\|x||$ for some $C \in (0,\infty)$ (and all $x$).
By hypothesis, the identity map from $(X,\|.\|_1)$ to $(X,\|.\|_2)$ is continuous. By Open Mapping Theorem its inverse is also continuous. This proves the existence of $c_1$ and $c_2$.